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Adding factors supported on the base can change the limiting alignment of a prime-denominator family. In base ten, the prime 53 gives two distinct subsequential limits, 80/689 and 82/689. Let b\ge2, let p\nmid b be prime, and let m be supported on the prime factors of b. We prove the exact formula \alpha_b(pm) = \frac{m - 1 + m T(C_m)/L}{pm - 1}. Here L=\operatorname{ord}_p(b), the reference tail follows the coset C_m=u\langle b\rangle with u=b^t/m and m\mid b^t, and T(C_m) records the digit-bin sizes encountered by that coset. The formula is exact at every finite m and uses one synchronized division depth for all fractions.
Different cosets can give different subsequential limits. If all cosets have the same normalized bin sum, as they do when b is a primitive root modulo p, the limit is (p-1+S)/(p(p-1)), where n_d counts the nonzero remainders emitting digit d and S=\sum_d n_d^2.
For every prime p\ge5 with p\nmid b, every coset, and every b-supported m, we prove the sharp bound \alpha_b(pm)<3/5<1/\varphi, where \varphi=(1+\sqrt5)/2. The constant 3/5 is the optimal uniform supremum. Among primes p\ge3 not dividing the base, p=3 is the only one that can cross the golden threshold. Denominators supported entirely on the base terminate and have alignment one by definition.
At p=53 in base ten, synchronized alignment does not settle to one value. An exhaustive enumeration of its four multiplicative cosets gives 80/689 for two cosets and 82/689 for the other two. The gap is small but exact. A single prime therefore carries two subsequential limits.
Splitting becomes possible when the digit function stops being injective. The base-supported part m creates a finite prefix, then chooses the multiplicative coset followed by the periodic reference tail. Once two remainder states can emit the same digit, different cosets can pass through digit bins of different sizes. The prime alone no longer determines the limit.
Hamming agreement between a reciprocal digit sequence and its cyclic shifts already appears in the communications literature [1]. There is also a mature arithmetic theory for digit values along a single multiplicative orbit. Girstmair relates full-period digit variance to Dedekind sums [3]. His work on class number factors studies sums over selected positions of a full period [2]. Murty and Thangadurai treat digit means when the base need not be a primitive root [4], and Girstmair develops mean and variance formulas for further proper orders [5]. The statistic here averages a digit-equality indicator over the complete denominator ensemble and isolates the remaining coset dependence in T(C_m). A primitive root gives one orbit containing every nonzero remainder, so no choice of coset remains. The contribution proved below is the exact coset formula together with the sharp uniform bound 3/5 across all orders and all cosets.
For digit-partitioning primes p\nmid b, characterized by p\le b+1, Digit-Partitioning Primes and the Alignment Formula [7] proves \alpha_b(pm)=\frac{2m-1}{pm-1}. Singleton digit bins make every coset indistinguishable in that range. Above the boundary, the same expression remains a lower bound, and the missing contribution depends on the coset selected by m.
The exact correction is carried by the coset bin sum T(C_m)=\sum_{r\in C_m}n_{\delta(r)}. It records where multiplication by the base meets the interval partition made by the floor function. The resulting formula covers every order and every coset. It also gives the sharp uniform bound 3/5 for every prime p\ge5 that does not divide the base, placing the entire range strictly below the golden threshold.
Definition 1 (Base-supported integers). An integer m\ge1 is b-supported if every prime factor of m divides b. Equivalently, m divides a power of b.
Definition 2 (Synchronized repetend alignment). Let p be prime, let m be b-supported, and put n=pm. If p\mid b, every fraction with denominator n terminates, and we set \alpha_b(n)=1. If p\nmid b, choose one depth t with m\mid b^t, and let L=\mathop{\mathrm{ord}}_p(b). Beginning immediately after the first t base-b digits, compare the next L digits of every k/n with the corresponding digits of 1/n. A terminating fraction receives score one; otherwise its score is its proportion of matching positions. The mean of these scores over 1\le k<n is the synchronized repetend alignment \alpha_b(n).
In the first branch, pm is itself b-supported. In the periodic branch, the definition is independent of the clearing depth. Increasing t advances both periodic tails by the same power of b, rotating both length-L words together without changing their match proportion. We never restart the individual fractions at separate canonical phases.
Definition 3 (Digit function and bins). For a prime p not dividing b, the digit function is \delta(r) = \lfloor br/p \rfloor for r \in \{0, \ldots, p{-}1\}. The bin of digit d is B_d = \{r \in \{1, \ldots, p{-}1\} : \delta(r) = d\}, and its size is n_d = |B_d|.
Lemma 4 (Digit-bin distribution). Write p-1=bq+r with 0\le r<b. Exactly r digit bins have size q+1, and the remaining b-r bins have size q. Therefore the sum of squared bin sizes is S(p,b) = \sum_{d=0}^{b-1} n_d^2 = bq^2 + r(2q + 1).
Proof. The condition \delta(x)=d places the integer x in the half-open interval \frac{dp}{b}\le x<\frac{(d+1)p}{b}. Since p is prime and p\nmid b, we have \gcd(p,b)=1 and 0<(r+1)/b<1. Each interval has length p/b=q+(r+1)/b, so each interior bin contains either q or q+1 integers. The first bin contains exactly the integers 1,\ldots,q, and reflection x\mapsto p-x gives the same count for the last bin. Thus every bin has size q or q+1. There are p-1=bq+r remainders altogether, so exactly r bins have the larger size. Hence S = r(q{+}1)^2 + (b{-}r)q^2 = rq^2 + 2rq + r + bq^2 - rq^2 = bq^2 + r(2q + 1). ◻
Corollary 5 (Singleton-bin square sum). When p \le b + 1, all bins have size at most 1, so S(p,b) = p - 1.
For n=pm with m a b-supported integer, choose t with m\mid b^t and set u=b^t/m. The coset C_m = u \langle b \rangle \bmod p is independent of the choice of t. If t'\ge t is another clearing depth, then b^{t'}/m=(b^t/m)b^{t'-t}, which lies in the same coset of \langle b\rangle. The reverse ordering follows by swapping the two depths. After the common prefix, every nonterminating row k/(pm) is governed by the nonzero residue uk\bmod p. In particular, the reference tail follows C_m, not necessarily the subgroup \langle b\rangle itself.
Definition 6 (Coset bin sum). For a coset C of \langle b \rangle in (\mathbb{Z}/p\mathbb{Z})^{*}, the coset bin sum is T(C) = \sum_{r \in C} n_{\delta(r)}, where n_d = |B_d| is the size of the d-th bin.
Different b-supported values of m can yield different cosets and therefore different bin sums.
Remark 7. The coset C has equidistributed bin sum if T(C)=\frac{LS}{p-1},\qquad S=\sum_d n_d^2. If every coset has this value, the alignment has one limit even when the selected coset changes with m. The stronger condition |C\cap B_d|=Ln_d/(p-1) for every d forces the displayed identity. When b is a primitive root modulo p, there is only one coset and the identity is automatic.
Theorem 8 (Exact coset alignment formula). For any prime p\nmid b and any b-supported m\ge1, the synchronized repetend alignment of n=pm is \alpha_b(pm) = \frac{m - 1 + \dfrac{mT(C_m)}{L}}{pm - 1}, where L=\mathop{\mathrm{ord}}_p(b) and C_m=u\langle b\rangle is determined by u=b^t/m with m\mid b^t.
Proof. The m-1 nonzero multiples of p below pm terminate and each contributes score one. The remaining (p-1)m numerators are nonterminating. After the common prefix, their starting states are uk\bmod p.
At synchronized position i, let r_i=b^iu\bmod p be the reference state. The state for k/(pm) is b^iuk\equiv kr_i\pmod p. As k ranges through the nonzero residue classes modulo p, the map k\mapsto kr_i is a permutation. Each residue class is represented by exactly m numerators below pm. Therefore the total match score at position i is m n_{\delta(r_i)}. Summing over one period and dividing by L, the nonterminating fractions contribute \frac{m}{L}\sum_{r\in C_m}n_{\delta(r)} =\frac{mT(C_m)}{L}. Adding the terminating contribution gives \alpha_b(pm)(pm-1)=(m-1)+\frac{mT(C_m)}{L}. ◻
Corollary 9 (The correction term). For every admissible p,b,m, \alpha_b(pm)-\frac{2m-1}{pm-1} =\frac{m\bigl(T(C_m)-L\bigr)}{L(pm-1)}\ge0. Equality holds exactly when every state in C_m lies in a singleton digit bin.
Proof. Every state belongs to a bin of size at least one, so T(C_m)\ge L. The identity follows by subtracting the two formulas. Equality means n_{\delta(r)}=1 for every r\in C_m. ◻
Remark 10. Along any sequence of b-supported values with m\to\infty for which the selected coset is fixed at C, the alignment tends to (L+T(C))/(pL). Different cosets can give different limits, so the unrestricted limit need not exist.
Remark 11. An exhaustive enumeration of all four cosets performed with nfield [9] for p = 53, b = 10, and L = 13 gives two pairs with distinct bin sums. The subgroup \langle 10 \rangle and one other coset have T = 67; the cosets 2\langle 10 \rangle and 5\langle 10 \rangle have T = 69. The two values give two subsequential limits, 80/689 and 82/689. For m = 2^a, u = 5^a, and the coset cycles with a \bmod 4.
Corollary 12 (Limit under equidistribution). Suppose every coset of \langle b\rangle has equidistributed bin sum in the sense of Remark 7. Then, as m\to\infty through b-supported integers, the alignment limit exists and equals \mathcal{L}(p,b) = \frac{p - 1 + S(p,b)}{p(p - 1)}, where, by Lemma 4, S(p,b) = bq^2 + r(2q{+}1) with q = \lfloor(p{-}1)/b\rfloor and r = (p{-}1) \bmod b. In particular, this holds whenever b is a primitive root modulo p.
Proof. For every coset, substitution gives \frac{L+T(C)}{pL} =\frac{L+LS/(p-1)}{pL} =\frac{p-1+S}{p(p-1)}. The value is independent of the coset, so it remains the limit even when C_m changes. If L=p-1, the unique coset contains every nonzero residue and T=S, so the hypothesis is automatic. ◻
Corollary 13 (Digit-partitioning recovery). For every digit-partitioning prime p\nmid b with p\le b+1 and every b-supported m, \alpha_b(pm)=\frac{2m-1}{pm-1}. Consequently, S=p-1 and \mathcal{L}(p,b)=2/p.
Proof. The digit function is injective, so every occupied bin is a singleton. Thus T(C_m)=L for every coset. Theorem 8 gives the finite formula, and its limit is 2/p. Corollary 5 gives S=p-1. ◻
The prime two is completely degenerate. If the base is even it belongs to the terminating case, while in an odd base the corollary gives \alpha_b(2m)=1. Thus it has alignment one in every base.
When b \equiv 1 \pmod p, the order L = 1 and \langle b \rangle = \{1\} is trivial. Every coset is a singleton, T(C_m) = n_{\delta(u)}, and for p \le b each singleton lies in its own bin with n_{\delta(u)}=1. The exact theorem then gives \alpha_b(pm) = \frac{(m{-}1) + m}{pm - 1} = \frac{2m-1}{pm-1}, recovering the union-count profile in Why the Golden Ratio Selects the Prime Three [6]. This is the one-position specialization of Corollary 13.
Corollary 12 gives the unique limit whenever all cosets have equidistributed bin sum.
For every displayed prime, nfield [9] exhausts every multiplicative coset and evaluates the values in Table 1.
| p | L | S | Equidist. | Subgroup limit | Decimal |
|---|---|---|---|---|---|
| 3 | 1 | 2 | 2/3 | 0.6667 | |
| 7 | 6 | 6 | 2/7 | 0.2857 | |
| 11 | 2 | 10 | 2/11 | 0.1818 | |
| 13 | 6 | 16 | 7/39 | 0.1795 | |
| 17 | 16 | 28 | 11/68 | 0.1618 | |
| 19 | 18 | 34 | 26/171 | 0.1520 | |
| 23 | 22 | 50 | 36/253 | 0.1423 | |
| 29 | 28 | 80 | 27/203 | 0.1330 | |
| 31 | 15 | 90 | 4/31 | 0.1290 | |
| 37 | 3 | 132 | 14/111 | 0.1261 | |
| 41 | 5 | 160 | 5/41 | 0.1220 | |
| 43 | 21 | 178 | 110/903 | 0.1218 | |
| 47 | 46 | 214 | 130/1081 | 0.1203 | |
| 53 | 13 | 272 | 80/689 | 0.1161 | |
| 59 | 58 | 338 | 198/1711 | 0.1157 | |
| 61 | 60 | 360 | 7/61 | 0.1148 |
Theorem 14 (Sharp uniform bound). For every prime p\ge5 with p\nmid b, every base b\ge2, and every b-supported m\ge1, \alpha_b(pm)<\frac35<\frac1\varphi. Every subsequential alignment limit is at most 3/5 and therefore remains strictly below 1/\varphi.
Proof. Put c=\lceil(p-1)/b\rceil. Every bin has size at most c, so T(C)/L\le c for every coset. The exact formula gives \alpha_b(pm) \le\frac{m-1+mc}{pm-1} <\frac{1+c}{p} \le\frac{p+1}{2p} \le\frac35. The strict step follows because the difference between its right and left sides is \frac{p-1-c}{p(pm-1)}>0. Indeed, b\ge2 and odd p\ge5 give c\le(p-1)/2<p-1. Finally, 3/5<1/\varphi because 3\varphi<5, equivalently 3\sqrt5<7. ◻
Proposition 15 (Sharpness). The constant 3/5 is the optimal uniform supremum.
Proof. Take p=5 and b=2. The base has order four modulo five, and both digit bins have size two. Hence T/L=2, so for m=2^a, \alpha_2(5m)=\frac{3m-1}{5m-1}\longrightarrow\frac35. Theorem 14 shows that no finite admissible value reaches or exceeds 3/5. ◻
Together, Theorem 14 and Proposition 15 give the best uniform bound. Thus among primes p\ge3 that do not divide the base, only p=3 can produce alignment above 1/\varphi. The result needs no equidistribution hypothesis and does not depend on the coset.
The quantity S(p,b)=\sum_d n_d^2 counts ordered pairs of nonzero remainders that emit the same digit. Therefore \frac{S(p,b)}{p-1} is the expected size of the bin containing a uniformly chosen remainder, while S(p,b)/(p-1)^2 is the collision probability for two independent uniform remainders.
An exhaustive nfield [9] enumeration of every coset for every prime below 100 that does not divide ten shows that all cosets have equidistributed bin sum except at 53, 73, 79, and 89. At these four primes, different cosets meet digit bins with different total sizes. A structural characterization of the pairs (p,b) for which every coset is equidistributed remains open. Character-sum methods provide the natural wider setting for that question [8].
For denominators pm with m supported on the base and p\nmid b prime, the exact formula accounts for every partial digit agreement. The example at 53 shows why the reference coset must be retained even as m grows. The sharp bound has a different source. No coset can encounter a bin larger than the floor partition allows, so every prime p\ge5 stays below 3/5.
Thus collisions can split the limiting alignment without allowing another prime p\ge3 to cross the golden threshold. A denominator whose part outside the base is composite requires a further argument; the prime formula does not classify that case.
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[5]K. Girstmair, Mean values and variances of the digits of 1/p, 2026. https://arxiv.org/abs/2606.29930.
[6]A. S. Petty, Why the Golden Ratio Selects the Prime Three, research note, January 2020 (revised September 2026), DOI: https://doi.org/10.5281/zenodo.20399951.
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[9]A. S. Petty, nfield, software repository. https://github.com/alexspetty/nfield
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